# Electric field acting on the source charge

Mayan Fung
I am reading Griffith's textbook on EM. There is a problem asking to find the force acting on the northern hemisphere by the southern hemisphere of a uniformly charged sphere.

The solution idea is to find the expression of the E field by Gauss's law and integrate the force over the northern hemisphere. However, part of the total electric field is contributed by the northern hemisphere itself. In my understanding, we should not include the electric field of the object we are calculating the force on it. I wonder why we can solve this problem with this approach?

Mentor
In my understanding, we should not include the electric field of the object we are calculating the force on it.
I think you are thinking of point particles.

• Vinh Nguyen
There is a straight-forward explanation if we break down the process of "calculating the force":

- In order to calculate the force acting on the Northern Hemisphere (NH), we can break the NH down into tiny fractions and calculate the "forces that act on each of those fractions", including:
--> The force that the Southern Hemisphere (SH) act on the fraction
--> The forces between NH fractions

- Then, we can calculate the force that act on the HM by adding all the "forces that act on each of those fractions". In doing so, we will have two sums:
--> Total force that the SH act on all the fractions of NH
===> This is the force that SH act NH as a whole
--> Total force of interaction between NH fractions
===> According to Newton's Third Law of Motion, every force has a counter-force, so the sum of the internal forces of all those fractions would ultimately add up to zero.

In short, we can claim that the force acting on the NH was from the SH (although there are internal interaction in the NH, those interactions canceled each other).

• starprince, Mayan Fung, vanhees71 and 1 other person
Mayan Fung
There is a straight-forward explanation if we break down the process of "calculating the force":

- In order to calculate the force acting on the Northern Hemisphere (NH), we can break the NH down into tiny fractions and calculate the "forces that act on each of those fractions", including:
--> The force that the Southern Hemisphere (SH) act on the fraction
--> The forces between NH fractions

- Then, we can calculate the force that act on the HM by adding all the "forces that act on each of those fractions". In doing so, we will have two sums:
--> Total force that the SH act on all the fractions of NH
===> This is the force that SH act NH as a whole
--> Total force of interaction between NH fractions
===> According to Newton's Third Law of Motion, every force has a counter-force, so the sum of the internal forces of all those fractions would ultimately add up to zero.

In short, we can claim that the force acting on the NH was from the SH (although there are internal interaction in the NH, those interactions canceled each other).

That's very clear. Thanks!

• • starprince, Dale and Vinh Nguyen